jee-main-electrostatics-important-questions-merged-pages-8

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Now 0AV
Q CV
x

  or 
0
Qx
V
A


 (ii) 
 Using (ii) in (i) we get 
 
2
0
Q
F
2 A


 
 Hence the correct choice is (c). 
12. The last three capacitors on the right, each of capacitance C = 9 F are in series and are equivalent 
to a capacitance C" given by 
 
1 1 1 1 1
or C' 3 F
C' 9 9 9 3
      . 
Since C is in parallel with C1, the equivalent capacitance of the last part of the network is C" = C' 
+ C1 = 3 + 6 = 9 F. Continuing this process of calculation towards the left, we notice that we are 
finally left with the combination whose equivalent capacitance is 3F. Hence the correct choice is 
(a). 
13. The three capacitors can be rearranged as shown in Fig. The capacitance between points P and S 
or between points Q and R = sum of the three capacitances = 3C = 9 F. Hence the correct choice 
is (d). 
 
14. C = 100 pF = 100 × 10
-12
 F = 10
-10
 F 
Let the number of sheets of foils required be n. They will form (n - 1) capacitors. If K is the 
dielectric constant of the dielectric, the capacitance is given 
 0K (n 1)A
C
d
 
 
 Or 
2
0 0
Cd Cd 4
n 1 .
K A K.4 r

  
  
 
 
2
0
4Cd
K.4 r


 
 
10 3 9
2 2
4 10 1 10 9 10
9
4 (1.0 10 )
 

    
 
 
 
 or n = 10 
Hence the correct choice is (a). 
15. If R is the radius of the big drop, we have 
 
 
 
3 34 R 4 r
1000
3 3
 
  
 which gives R = 10 r. The electrical potential of each droplet is 
 
0
q
v
4 r


 
 and that of the big drop is 
 
0
1000q
V
4 R


 
  
V 1000r
100
v R
  (R = 10r) 
 Hence the correct choice is (c). 
16. If A is the area of each plate, the capacitance of the air capacitor shown in Fig.(a) is 
 0
0
A
C
d

 , where C0 = 2 F (given). 
 The capacitance of air capacitor in Fig. (b) is 
 0
1 0
A 2 A
C 2C
d / 2 d
 
   
 The capacitance of the dielectric filled capacitor in Fig. (b) is 
 0
2 0
k A 2k A
C 2kC
d / 2 d
 
   
where k is the dielectric constant. Now capacitors C1 and C2 are in series. Therefore, the 
capacitance C of the capacitor shown in Fig. (b) is given by 
 
1 2 0 0 0
1 1 1 1 1 (k 1)
C C C 2C 2kC 2C k

     
 Or 02C k 2 2 F 3
C 3 F
(k 1) (3 1)
  
   
 
 
 Hence the correct choice is (c). 
17. If A is the area of each plate, the capacitance of the air-filled capacitor shown in Fig. (a) is 
 0
0
A
C ,
d

 where C0 = 2F (given). 
 The capacitance of air capacitor in Fig. (b) is 
 0 0 0
1
A / 2 A C
C
d 2d 2
 
   
 The capacitance of dielectric filled capacitor in Fig. (b)is 
 0 0 0
2
k A / 2 k A kC
C
d 2d 2
 
   
 Since C1 and C2 are in parallel, the capacitance C of the capacitor shown in Fig. (b) is 
 
 
 0 0
1 2
C kC
C C C
2 2
    
 0C 2 F
(1 k) (1 3) 4 F
2 2

      
 Hence the correct choice is (a). 
18. Refer to Fig. The charge at Q exerts an attractive force F on charge at P along PQ. The charge at R 
exerts a repulsive on charge at P along PS of magnitude F. The angle between these two forces is 
120°. From parallelogram law, the magnitude of the resultant force is 
 F
2
1 = F
2 
+ F
2
 + 2F
2
 cos 120° = 2F
2
 - F
2
 = F
2 
 
or Fr = F. As shown in the figure, the direction of the resultant force is along the negative x- 
direction. Hence the correct choice is (b). 
19. Refer to Fig. Charges at P and R both exert an attractive force on charge at Q. The angle between 
these forces is 60°. The resultant force has a magnitude 
 
 F
2
 = F
2
 + F
2
 + 2F
2
 cos 60° 
 = 2F
2
 + F
2
 = 3F
2 
 
 Fr = 3F 
The angle  is given by 
 
o
o
Fsin 60 1
tan
F Fcos60 3
  

 
   = 30°. 
 Hence the correct choice is (d). 
20. Refer to Fig. Let us consider forces on a ball, say, Q. Three forces act on it: (i) tension T in the 
thread, (ii) force mg due to gravity and (iii) force F due to Coulomb repulsion along +ve x-
direction. For equilibrium, the sum of the x and y components of these forces must be zero, i.e. 
 T cos 60° - F= 0 
 
 
 and T sin 60° - mg= 0 
 These equations give F = mg cot 60° = 
3
10
× 10
-3
 × 10 × 
1
3
 = 10
-3
 N. Now 
 
 
2
2
0
1 q
F .
4 r


 
 Putting F = 10
-3
 N, r = 0.3 m and 
0
1
4
= 9 × 10
9
, we get q = 10
-7
 coulomb. 
21. Refer to Fig. The electric field E1 at (a, b) due to q1 has a magnitude 
 1
1 2
0
q1
E .
4 a


 
 
 and is directed along + x-axis. The electric field E2 at (a, b) due to q2 has a magnitude 
 2
2 2
0
q1
E .
4 b


 
and is directed along +y-axis. The angel  subtended by the resultant field E with the x-axis is 
given by 
 
22
2 2
2
1 1
E q a 1 2
tan . 2
E q b 2 1
 
      
 
 
 Hence the correct choice is (b). 
22. The correct choice is (d). The electric field E exerts a force qE on charge + q and a force - qE on 
charge - q of the dipole. Since these forces are equal and opposite, they add upto zero. 
23. The correct choice is (b). A torque acts on the dipole which tends to align it along the field. 
24. The correct choice is (c). In a non-uniform electric field, a dipole experiences a force which gives 
it a translational motion and a torque which gives it a rotational motion. 
25. Refer to Fig. Potential at O is 
 
 
 0
0
1 q q q q
V 0
4 r r r r
 
     
  
 
 Hence the correct choice is (d). 
 
26. The distance of a vertex from the centre of the cube of side b is r = 3 b/2. Now the potential due 
to charge q at the centre is q/4 0r. Hence the potential due to the arrangement of eight charges 
(each of magnitude q) at the centre is 
 
0 0
8q 4q
V
4 r 3 b
 
 
 
27. We know that electric fields are to be added vectorially. From the symmetry of the eight charges 
with respect to the centre of the cube, it is evident that the electric fields at the centre due to two 
opposite charges cancel in pairs (being equal and opposite). Hence the net electric field at the 
centre of the cube will be zero. 
28. Refer to Fig. The distance of point P1 from charge + q is r1 = z - a and from charge - q is 
 r2 = z + a. 
 
  Potential at Pl = 
0 1 2
1 q q
4 r r
 
 
  
 
 2 1
0 1 2
r rq
.
4 r r



 
 
2 2
0
2qa
,
4 (z a )

 
 
 which is choice (c). 
29. Refer to Fig. again. Any point on the perpendicular bisector passing through the centre of the 
dipole is at the same distance from the two charges. Hence the potentials at point P2(5, 0, 0) and 
that at point P3 (- 7, 0, 0) are zero. Since P2 and P3 are at the same potential (zero), the potential 
 
 
difference between them is zero. Hence no work will be done in moving a charge from P2 to P3, 
The answer will not change if the path of the charge is changed because the work done is 
independent of the path taken. 
30. Refer to Fig. The total potential energy of the arrangement of charges is the sum of the energies of 
each pair of charges. The potential energy of the system comprising the three charges q1, q2 and q3 
is 
 
 U = W1 + W2 + W3 
 1 3 2 31 2
0 12 13 23
q q q qq q1
4 r r r
 
   
  
 (i) 
Here q1 = q2 = q = + 1.6 × 10
-19
 C (proton), q3 = -q = - 1.6 × 10
-19
 C (electron), r12 = 1.5 Å = 1.5 × 
10
-10 
m,
 
r13 = r23 = 1 Å = 1 × 10
-10
 m and 1/4 n0 = 9 × 10
9 
Nm
2
C
-2
. Thus 
 
2 10
0
4 q 10
U . joule
3 4
 


 
 
10
0
4 q 10
. eV
3 4

 

 
 
19 10 94 1.6 10 10 9 10
3
    
  
 = 19.2eV 
31. Charge on electron (- e) = - 1.6 × 10
-19
 C, charge on proton (e) = 1.6 × 10
-19
 C, separation r = 0.53 
Å = 0.53 × 10
-10
 m. If the zero of potential energy is taken to be at infinite separation, the potential 
energy of the electron-proton system is 
 
2
0
1 e
U . joule
4 r
 

 
 
0
1 e
. eV
4 r
 

 
 
9 19
10
9 10 (1.6 10 )
27.2eV
0.53 10


  
  
Hence the correct choice is (b). 
32. If the electron was at rest, 27.2 eV of the energy will have to be supplied (or 27.2 × 1.6 × 10
-19
 J of 
work will have to be done) to free the electron from the attraction of the proton and remove it to 
infinity. Since the electron is moving (round the proton) with a kinetic energy = U/2 = 1/2 × (- 
 
 
27.2) = - 13.6 eV, the electron itself is supplying an energy of 13.6 eV due to centrifugal action. 
Hence the minimum amount of work required to free the electron = 27.2 - 13.6 = 13.6 eV = 13.6 × 
1.6 × 10
-19
 = 2.2 × 10
-18
 J. Hence the correct choice is (d). 
33. The potential energy of electron-proton system at a separation of 1.06 Å = half that at a separation 
of 0.53 Å = half of - 27.2 eV = - 13.6 eV. If the zero of potential energy at a separation of 1.06 Å 
is taken to be zero (instead of- 13.6 eV), the potential energy of the electron-proton system would 
be 
 = - 27.2 - (- 13.6) = - 13.6 eV, which is choice (b). 
34. Since the potential energy of the system is now - 13.6 eV, the energy supplied by the electron 
itself is 13.6 eV by virtue of its orbital motion round the proton. Hence the minimum work to pull 
the electron from the atom will be zero. 
35. The series combination of C2 and C3 is equivalent to a capacitance C' given by 
 
2 3
1 1 1
C' C C
  
 Or 2 3
2 3
C C 200 200
C' 100pF
C C 200 200

  
 
 
 
Therefore the circuit reduces to the one shown in Fig. (a). The equivalent capacitance between 
points A and B is 
 C" = C1 + C = 100 + 100 = 200 pF 
The circuit may be further simplified to that in Fig. (b). The equivalent capacitance C of the entire 
network, i.e., between points A and D, is now that of the series combination of C" and C4. Thus 
 
4
1 1 1 1 1
C C" C 200 100
    
 
3 200
or C pF
200 3
  
 Hence the correct choice is (c). 
36. The series combination of 6 and 12 is equivalent to 4 and the parallel combination of 2 and 2 is 
also equivalent to 4. Therefore the network can be simplified as shown in Fig. 
The parallel combination of 4 and 4 is equivalent to 8 and the series combination of 8 and 4 is 
equivalent to 8/3. Thus the combination in Fig.1 reduces to that in Fig.2 
 The series combination of 1 and 8 in Fig.2 yields 8/9 as shown in Fig. 3